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Let a fixed simple three-uniform hypergraph $F$ have $m$ hypergraph edges, with one host vertex of a hypergraph class $V_i$ for each vertex of a hypergraph. Assume each hypergraph edge triple has indicator $h_e$, subset density $p_e$, and balanced three-dimensional box norm at most ${\alpha }^{1/8}$. The normalized labelled hypergraph copy count is $\mathbb{E}{\prod }_{e\in E(F)}{h}_e$. Telescope this product against ${\prod }_e{p}_e$. In each error term, fix all variables outside the three vertices of a hypergraph of its balanced hypergraph edge. Every other hypergraph edge intersects that triple in at most two vertices of a hypergraph, so the remaining factors group into three pair functions bounded by one. The pair-factor correlation bound for the three-dimensional box norm bounds the error by ${\alpha }^{1/8}$. Summing givesFor the three-uniform tetrahedron, $m=4$. The same argument with a common host set counts homomorphisms; excluding collisions of distinct abstract vertices of a hypergraph changes the count by at most $\left(\genfrac{}{}{0ex}{}{|V(F)|}{2}\right)|V{|}^{|V(F)|-1}$.

For hypergraphs $F,H$, a labelled copy is an injective function $\varphi :V(F)\to V(H)$ taking every hypergraph edge of $F$ to an hypergraph edge of $H$. Specified vertex of a hypergraph classes may additionally require $\varphi (i)\in V_i$. The count is the sum, over these injective maps, of the product of the host hypergraph edge indicators. Without injectivity the same expression counts hypergraph homomorphisms. On a common host of size $n$, their counts differ by at most $\left(\genfrac{}{}{0ex}{}{|V(F)|}{2}\right)n^{|V(F)|-1}$: every noninjective map identifies at least one pair of abstract vertices of a hypergraph, each prescribed equality permits at most $n^{|V(F)|-1}$ maps, and the union bound completes the estimate. An induced copy additionally preserves nonedges, so its indicator product includes nonedge factors. The counting lemma for octahedrally quasirandom three-uniform hypergraphs estimates these labelled products when the requisite centered triple indicators have small box norm.

A hypergraph is linear when two distinct hypergraph edges intersect in at most one vertex of a hypergraph. Equivalently, every pair of distinct vertices of a hypergraph has hypergraph codegree at most one. This restricts local overlap without bounding the hypergraph chromatic number: the random alteration method gives linear three-uniform hypergraphs of arbitrarily large hypergraph chromatic number.

A multipartite hypergraph has a specified partition of its vertex of a hypergraph set into parts, and every hypergraph edge meets each part in at most one vertex of a hypergraph. For an $r$-uniform hypergraph with $r$ parts, an hypergraph edge chooses exactly one vertex of a hypergraph from each part; its indicator is therefore a function on the Cartesian product of those parts. A quadripartite three-uniform hypergraph instead has four possible types of hypergraph edge, according to the omitted part. A choice of one vertex of a hypergraph from each of its four parts is a three-uniform tetrahedron exactly when all four triples are hypergraph edges. Specifying the parts makes both subset density and labelled counting precise.

Yes. The same telescoping argument proves the counting lemma for octahedrally quasirandom three-uniform hypergraphs for every fixed simple three-uniform hypergraph $F$.

Label its vertices of a hypergraph $1,\dots ,v$ and give vertex of a hypergraph $i$ a nonempty host part $V_i$. For each hypergraph edge $e=\{i,j,k\}$, let $h_e$ be the corresponding host hypergraph edge indicator function of subset density $p_e$, with $\Vert h_e-p_e{\Vert }_{{\square }^3}\le {\alpha }^{1/8}$. The normalized labelled hypergraph copy count isIf the host parts are disjoint, every such vertex of a hypergraph choice is automatically injective. Enumerate the $m$ hypergraph edges as $e_1,\dots ,e_m$ and telescopeIn the term for $e_j=\{a,b,c\}$, fix all variables outside this triple. Since $F$ is simple and three-uniform, every other hypergraph edge intersects $\{a,b,c\}$ in at most two vertices of a hypergraph. Its remaining factor therefore depends on at most two of $x_a,x_b,x_c$. Group all two-variable factors according to the pairs $(a,b),(a,c),(b,c)$. Absorb each one-variable factor into any pair containing its variable, and constants into any group. The three resulting pair functions remain bounded by one.

The balanced factor is now multiplied by exactly the sort of pair functions handled in part (ii). Its average has absolute value at most ${\alpha }^{1/8}$, independently of the fixed outside variables. Averaging those variables and summing the $m$ errors givesEquivalently,This counts labelled copies preserving the specified vertex of a hypergraph classes, for arbitrary intersection patterns among the hypergraph edges, and includes the three-uniform tetrahedron as a special case.

For a common host vertex of a hypergraph set of size $n$, the same product average counts homomorphisms. At most $\left(\genfrac{}{}{0ex}{}{v}{2}\right)n^{v-1}$ assignments identify a pair of distinct labelled vertices of a hypergraph, by the union bound. Thus restricting to injective copies changes the estimate by at most this amount. With a common normalized subset density $p$ the injective count is therefore $p^m{n}^v+O(m{\alpha }^{1/8}{n}^v+v^2{n}^{v-1})$. For induced copies, include the nonhypergraph edge indicator functions as additional factors; their balanced functions are negatives of the corresponding edge-balanced functions and have the same box norm. The same proof applies provided all the relevant triple parts satisfy the stated quasirandomness condition.

First prove the pair-factor correlation bound for the three-dimensional box norm. For real $g$ and three functions $a(y,z),b(x,z),c(x,y)$ bounded in absolute value by one, letApply the Cauchy-Schwarz inequality in $(y,z)$, placing the $x$ average inside and removing $a$. This givesExpand this square, with independent $x_0,x_1$. Apply Cauchy-Schwarz in $(x_0,x_1,z)$, removing the product $b(x_0,z)b(x_1,z)$. The result isExpand with independent $y_0,y_1$ and apply Cauchy-Schwarz in $(x_0,x_1,y_0,y_1)$, removing the four $c$ factors. ThenTherefore $|T|\le \Vert g{\Vert }_{{\square }^3}$, with no independence assumption on the three pair functions.

In this multipartite hypergraph, a simplex is a three-uniform tetrahedron: one vertex of a hypergraph from each of $X,Y,Z,W$, with all four possible triples present. Write the corresponding hypergraph edge indicator functions as $h_1(x,y,z),h_2(x,y,w),h_3(x,z,w),h_4(y,z,w)$, of subset densities $p,q,r,s$. Its normalized count isTelescope the product exactly:For the first term fix $w$. The remaining three factors are bounded pair functions of $(x,y)$, $(x,z)$ and $(y,z)$, so the inequality just proved bounds its $(x,y,z)$ average by $\Vert h_1-p{\Vert }_{{\square }^3}\le {\alpha }^{1/8}$. Averaging over $w$ preserves that bound. For each other term fix the vertex of a hypergraph outside its balanced triple and apply the same argument, allowing an unused pair factor to be identically one. The subset density prefactors are at most one. HenceMultiplying by $|X||Y||Z||W|$ givesThus one may take the absolute constant $C=4$. This proves the complete-pair-support case of the tetrahedron counting lemma directly; no lower bound on $p,q,r,s$ is needed.